hdu 1003 Max sum

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 208607    Accepted Submission(s): 48835

Problem Description

Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max sum of a sub-sequence. For example, given (6,-1,5,4,-7), the max sum in this sequence is 6 + (-1) + 5 + 4 = 14.

 

Input

The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line starts with a number N(1<=N<=100000), then N integers followed(all the integers are between -1000 and 1000).

 

Output

For each test case, you should output two lines. The first line is "Case #:", # means the number of the test case. The second line contains three integers, the Max Sum in the sequence, the start position of the sub-sequence, the end position of the sub-sequence. If there are more than one result, output the first one. Output a blank line between two cases.

 

Sample Input

2 5 6 -1 5 4 -7 7 0 6 -1 1 -6 7 -5

 

Sample Output

Case 1: 14 1 4 Case 2: 7 1 6

#include <iostream>
#define N 100010
using namespace std;
int a[N], d[N];
int main()
{
    int test, n, i, max, k, b, e;
    cin >> test;
    k =1;

    while (test--)
    {
        cin >> n;
        for ( i = 1; i <= n; i++)
        {
            cin >> a[i];
        }
        d[1] = a[1];
        for ( i = 2; i <= n; i++)
        {
            d[i] = (d[i-1] + a[i] > a[i] ? d[i-1] + a[i] : a[i]);
        }
        max = d[1];
        e = 1;
        for (i = 2; i <= n; i++)
        {
            if(d[i] > max){
                max = d[i];
                e = i;
            }
        }
        int t = 0;
        b = e;
        for (i = e; i > 0; i--)
        {
            t += a[i];
            if (t == max){
                b = i;
            }
        }
        cout << "Case " << k++ << ":" << endl << max << " " << b << " " << e << endl;
        if(test)
            cout << endl;
    }
    return 0;
}